# Why in Overlap Save Method in DSP output length comes out to be same as input length?

I know that when we do convolution of $x(n)$ (Length $L$) and $h(m)$ (Length=$M$), length of output data comes out to be $L+M-1$, but when we break a large sequence of data and find output using Overlap Save Method, length comes out to be same as input.

• Why do the extra $M-1$ points are appearing in this case?

In more expressive way, be $y(n) = x(n)*h(n)$:

Suppose Case 1.:

• $x(n)$ (length $L$),
• $h(n)$ (length $M$), leads to
• $y(n)$ (length $L+M-1$)

using simple linear convolution

Case 2.:

• $x(n)$ (length $3L$),
• $h(n)$ (length $M$), leads to
• $y(n)$ (length $3L$)

using the Overlap Save Method that evaluats output after breaking input in 3 equal parts of length $L$.

• It is not clear to me: Have you understood how the Overlap-Save method works? If you are fully understanding how Overlap-Save works, where does the question come from? – Marcus Müller Feb 19 '17 at 18:01
• Also, I had to completely restructure your question just to be able to read it. Please make sure your question is nicely structured yourself the next time. – Marcus Müller Feb 19 '17 at 18:03
• Are processors use different techniques (other than linear de-convolution) to interpret the input signal from the output response of systems.If answer is no then why output (having length L+M-1) evaluted using linear convolution is different from the output (having length same as input signal) given by overlap-save method. – Vikash Yadav Feb 19 '17 at 20:37
• hotpaw2's answer explains that. You did not understand the overlap-save method if you're still asking this. – Marcus Müller Feb 19 '17 at 20:53